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Practical Design to Eurocode 2
Foundations
Eurocode 7
Eurocode 7 has two parts:
Part 1: General Rules
Part 2: Ground Investigation and testing
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Limit States
The following ultimate limit states apply to foundation
design:
EQU: Loss of equilibrium of the structure
STR: Internal failure or excessive deformation of thestructure or structural member
GEO: Failure due to excessive deformation of the ground
UPL: Loss of equilibrium due to uplift by water pressure
HYD: Failure caused by hydraulic gradients
Categories of Structures
Large or
unusual
structures
Abnormal risksAll other structures3
Spread
foundations
No exceptional
risk
Conventional types of
structure no difficult
ground
2
None givenNegligibleSmall and relatively
simple structures
1
Examples from
EC7
Risk of
geotechnical
failure
DescriptionCategory
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STR/GEO ULS
1.30,iQk1.3Qk1.0Gk1.0GkExp 6.10
Notes:
If the variation in permanent action is significant, use Gk,j,sup and Gk,j,infIf the action if favourable, Q,i = 0 and the variable actions should be ignored
Combination 2
1.50,iQk1.5Qk1.0Gk1.25GkExp 6.10b
1.50,iQk1.50,1Qk1.0Gk1.35GkExp 6.10a
1.50,iQk1.5Qk1.0Gk1.35GkExp 6.10
Combination 1
OthersMainFavourableUnfavourable
Accompanying variableactions
Leadingvariableaction
Permanent Actions
Partial factors
1.01.0Bulk density
1.41.0quUnconfined strength
1.41.0cuUndrained shear
strength
1.251.0cEffective cohesion
1.251.0Angle of shearing
resistance
Combination 2Combination 1Symbol
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Spread Foundations
EC7 Section 6
Three methods for design:
Direct method check all limit states
Indirect method experience and testing used to
determine SLS parameters that also satisfy ULS
Prescriptive methods use presumed bearing
resistance (BS8004 quoted in NA)
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Pressure distributions
EQU : 0.9 Gk + 1.5 Qk (assuming variable action is
destabilising e.g. wind, and permanent action is
stabilizing)
STR : 1.35 Gk + 1.5 Qk (6.10)
(6.10a or 6.10b could be used)
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a a
bF
hF
(9gd/fctd,pl)h
a
F0,85
gd is the design value of the ground pressure
as a simplification hf/a 2 may be used
Strip and Pad Footings(12.9.3) Plain concrete
Reinforced Bases
Check critical bending moment at column face
Check beam shear and punching shear
For punching shear
the ground reaction
within the perimeter
may be deducted
from the column load
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Worked Example
Design a square pad footing for a 350 350 mm columncarrying Gk = 600 kN and Qk = 505 kN. The presumed
allowable bearing pressure of the non-aggressive soil is
200 kN/m2.
Category 2, using prescriptive methods
Base size: (600 + 505)/200 = 5.525m2
=> 2.4 x 2.4 base x .5m (say) deep.
Worked Example
Use C30/37
Loading = 1.35 x 600 + 1.5 x 505
= 1567.5kN
ULS bearing pressure =1567.5/2.42
= 272kN/m2
Critical section at face of column
MEd = 2.72 x 2.4 x 1.0252 / 2
= 343kNm
d = 500 50 16 = 434mm
K = 343 x 106 / (2400 x 4342 x 30)
= 0.025
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Worked Example
z = 0.95d = 0.95 x 434 = 412mm
As = MEd/fydz = 343 x 106 / (435 x 412) = 1914mm2
Provide H16 @ 250 c/c (1930mm2)
Beam shear
Check critical section d away from column face
VEd = 272 x (1.025 0.434) = 161kN/m
vEd = 161 / 434 = 0.37MPa
vRd,c (from table) = 0.41MPa => beam shear ok.
Worked Example
Punching shear
Basic control perimeter at 2d from face of column
vEd = VEd / uid < vRd,c
= 1, ui = (350 x 4 + 434 x 2 x 2 x ) = 6854mm
VEd = load minus net upward force within the area of the
control perimeter)
= 1567.5 272 x (0.352 + x .8682 + .868 x .35 x 4)
= 560kN
vEd = 0.188MPa; vRd,c = 0.41 (as before) => ok
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Workshop Example
Pad foundation for a column taking Gk = 300kN, Qk =
160kN. Permissible bearing stress = 150kPa.
Work out size of base, tension reinforcement and any
shear reinforcement.
Retaining Walls
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Ultimate Limit States
for the design ofretaining walls
General expressions
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Calculation Model A
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Calculation Model B
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Partial factors
1.01.0Bulk density
1.41.0quUnconfined strength
1.41.0cuUndrained shear
strength
1.251.0cEffective cohesion
1.251.0Angle of shearing
resistance
Combination 2Combination 1Symbol
Overall design
procedure
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Initial sizing
Figure 6 for overall
design procedure
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Panel 2
Overall design
procedure
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Design against
sliding
(Figure 7)
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Overall design
procedure
Design against Toppling(Figure 9)
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Overall design
procedure
Design against bearing failure(Figure 10)
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Overall design
procedure
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Figure 13
Strut and tie methods
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P1
P2 2P1
Which is the Stronger?
fcu
fcu
fct
fct
compressive strength of
concrete with transverse
tension
tensile stress in concrete
Bi-axial Strength of Concrete
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EC2 provides a simplified approach to limiting stresses in struts.
Where there is no transverse tension
Rd,max =fcd (ie 0,57fck)
Otherwise where there is transverse tension Rd,max = 0,6fcd (ieabout 0,3fck)
where = 1-fck/250
Struts
Strut and Tie Models(6.5.2)
Sizing nodes (6.5.4)
The size of a node is
determined by the limiting
stresses in the struts and the
anchorage length of bars
If the stresses in all struts
meeting at a node are the
same, the boundaries of the
node will be perpendicular to
the axis of the strut and the
node will be in a state ofhydrostatic stress
Bars can be
anchored boththrough and beyond
a node
c-c-c(1-fck/250)fcd
(Exp (6.60))
Unless special confinement is
provided, the calculated
compressive stress in the node
regions should not exceed:
c-c-t0.85(1-fck/250)fcd
(Exp (6.61))
c-t-t0.75(1-fck/250)fcd
(Exp (6.62))
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1 - Anchorage of tension reinforcement at simple support
Even at a simple support there is tension
in the reinforcement
behaviour similar to deep beam at support
Strut and Tie Models Can HelpUs Understand .
strut needs tobe supported
on bar
bar to be anchored
beyond point of load
application
2 - Detailing of nib
bearing
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3 - Anchorage of bars
F
F/2 F/2
F tan
F tan
Forces from cranks reduced if laps in plane parallel to surface
but similar splitting forces from lap
F F
4 - Lapping of bars
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5.5 MN (Ult)
7503400
1650
fck = 30 MPa / fyk = 500 MPa
breadth = 1050
bottom cover = 100, bars 32mm
side cover = 75
1. Set up an arrangement of
struts and ties
2. Check the strength of the
struts 6.5.2 (2)
3. Check the strength of thenodes 6.5.4 (4)
4. Calculate the areas of
reinforcement required
5. What other checks should
be made to EC2?
700 x 700 column
200
Pile Cap Exercise
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1. Arrangement of struts andties
Compressive force in eachstrut
= 2750 / sin 53.7
= 3412 kN
Tensile force in tie
= 2750 / tan 53.7
= 2020 kN
Width of each strut
= 700 sin 53.7
= 564 mm
Angle of strut
= tan -1 (1530 / 1125)
= 53.7
1125
2750 kN
1530
120
53.7
5500 kN
700 x 700 column
2750 kN
2020 kN
3412 kN
564
Pile Cap Exercise Solution (1)
2. Strength of struts
Stress in strut:Projecting the rectangular geometryfrom the column gives minimum area,
and maximum stress= 3.412 / (0.564 x 0.7)= 8.6 MPa
Cl 6.5.2 (2) : Strength of strut
sRd,max = 0.6 fcd
= 0.6 (1-fck/250)fcd
= 0.6 (1 30 / 250)0.85 x 30 / 1.5
= 8.98 MPa OK
53.7
3412 kN
564
700
Pile Cap Exercise Solution (2)
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Upper node:
Stress on hor. plane Ed,1= 2.75 / (0.35 x 0.7)= 11.2 MPa
Stress on vert. plane Ed,2= 2.02 / (0.35 tan 53.7 x 0.7)= 6.1 MPa
As these will be principle stresses (noshear), the other plane is ok by inspection.
Cl 6.5.4 (4a) :Strength of compression node without ties= Rd,max= (1-fck/250)fcd= (1 30 / 250) 0.85 x 30 / 1.5
= 14.96 MPa OK
3. Strength of nodes
3412 kN
700mm
2.02MN
Ed,2
Ed,1
5.5 MN (Ult)
Pile Cap Exercise Solution (3)
3. Strength of nodes
Lower node:The exact geometry of node is difficult todefine, so conservatively compare strutstresses to node allowable stresses
Stress in strut Ed,1= 8.6 MPa
Stress in pile Ed,2= 2.75 / (0.752 / 4)= 6.2 MPa
Cl 6.5.4 (4b) :
Strength of compression node with anchored ties
= Rd,max= 0.85 (1-f
ck/250)f
cd
= 0.85 (1 30 / 250) 0.85 x 30 / 1.5
= 12.7 MPa OK
2750 kN2750 kN
Ed,2
Ed,2
750 750
Ed,1Ed,1
Pile Cap Exercise Solution (4)
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4. Area of reinforcement
Tension = 2020 kN
Area of steel required
= 2020 x 1000/ (500/1.15)
= 4646 mm2
6H32 (4825 mm2)
5. Other checks
Assume bar fully stressed at centreline of pile
Standard mandrel diameter (Cl 8.3 (3) & Table 8.1N)m = 7 = 7 x 32 = 224 mmStandard extension of bar beyond bend = 5 = 5 x 32 = 160 mm
Cl 8.4.4. (1) :1
=0.7;, 2
= (1-0.15(91-32)/32 = 0.72;3
= 1;4
= 1;
5 = 1 0.04 (2.75/(1.5x.75)) = 0.902; so 235 = 0.649 but Exp(8.5) 235 = 0.7
Hence lbd = 12 345Lb,rqd = 0.7 x 0.7 x 1116.3 = 547 mm
53.7
2750 / tan53.7= 2020 kN
2750 kN
112 mm
160mm
Anchorage length, lbd :
Cl 8.4.2 (2) :fbd= 2.25 12fctd= 2.25 x 1 x 1 x 1x 2/1.5 = 3.0 MPa
Cl 8.4.3 (2) : lb,rqd= (/4)(sd/fbd)
= (32/4) x {2020,000 x 4/(6 x x322 x 3)}= 1116.3 mm
Pile Cap Exercise Solution (5)
Tying
www.eurocode2.info
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Peripheral ties (9.10.2.2):
Ftie,per = (20 + 4n0) 60kN where n0 is the number of storeys
Internal ties (including transverse ties) (9.10.2.3):
Ftie,int = ((gk + qk)/7.5)(lr/5)Ft Ft kN/m
Where (gk + qk) is the sum of the average permanent and variable floor
loads (kN/m2), lr is the greater of the distances (m) between the centres of
the columns, frames or walls supporting any two adjacent floor spans in the
direction of the tie under consideration and
Ft = (20 + 4n0) 60kN.Maximum spacing of internal ties = 1.5 lr
Horizontal ties to columns or walls (9.10.2.4):Ftie,fac = Ftie,col (2 Ft (ls /2.5)Ft) and 3% of the total design ultimate verticalload carried by the column or wall at that level. Tying of external walls is only
required if the peripheral tie is not located within the wall. Ftie,fac in kN per
metre run of wall, Ftie,col in kN per column and ls is the floor to ceiling height in
m.
Tying systems (2)
Vertical ties (9.10.2.5):In panel buildings of 5 storeys or more, ties should be provided in
columns and/or walls to limit damage of collapse of a floor.
Normally continuous vertical ties should be provided from the
lowest to the highest level.
Where a column or wall is supported at the bottom by a beam or
slab accidental loss of this element should be considered.
Continuity and anchorage ties (9.10.3):Ties in two horizontal directions shall be effectively continuous andanchored at the perimeter of the structure.
Ties may be provided wholly in the insitu concrete topping or at
connections of precast members.
Tying systems (3)